Symmetry assignment of spin-flip states in EOM

A fundamental question for the EOM experts…

The Q-Chem manual notes the following:

It is a symmetry of a transition rather than that of a target state which is specified in excited state calculations. The symmetry of the target state is a product of the symmetry of the reference state and the transition. For closed-shell molecules, the former is fully symmetric and the symmetry of the target state is the same as that of transition, however, for open-shell references this is not so.

Does this same consideration apply to spin-flip EOM calculations? Or can the symmetry of the SF states be directly read off of the output?


The EOM code always prints the symmetry of the transition, so to get symmetry of the target state, you should figure out the symmetry of the reference. The easiest way to do that is to look at the output in the beginning of the CCMAN2 calculation:

Point group: C2v (4 irreducible representations).

                       A1   A2   B1   B2    All 

All molecular orbitals:

  • Alpha 7 0 4 2 13
  • Beta 7 0 4 2 13

Alpha orbitals:

  • Frozen occupied 1 0 0 0 1
  • Active occupied 2 0 1 1 4
  • Active virtual 4 0 3 1 8
  • Frozen virtual 0 0 0 0 0

Beta orbitals:

  • Frozen occupied 1 0 0 0 1
  • Active occupied 1 0 1 0 2
  • Active virtual 5 0 3 2 10
  • Frozen virtual 0 0 0 0 0

From this you can see that your SOCC are a1 and b2, so your reference triplet is B2 symmetry. Hence, this means that SF transitions of A1 symmetry will give you B2 states and transitions of B2 symmetry will give you A1 states.

Here is the link to the materials from our most recent workshop:
Google drive with workshop materials
Check out CC/EOM folder and look at the pdf file presenting the EOM-SF calculations and the analysis. Inputs and outputs are also there.
The recordings are available here:

Hope this helps – and sorry for the slow response.

Hi, Anna. A very belated thank-you here. This info was very helpful!